Read the passage carefully and answer the Questions.
\( KMnO_4 \) is prepared by the fusion of \( MnO_2 \) with an alkali metal hydroxide and an oxidizing agent like \( KNO_3 \) to give a dark- green manganate ion which disproportionates to give permanganate as follows.
\( 2MnO_2+ 4KOH + O_2→2K_2MnO_4+2H_2O \)
\( 3K_2MnO_4+ 4H^+→2KMnO_4 + MnO_2 + 2H_2O \)
On heating \( KMnO_4 \) decomposes at 513 K to give \( K_2MnO_4 \). Permanganate ion is tetrahedral and diamagnetic. Acidified \( KMnO_4 \) acts as a strong oxidizing agent which oxidizes oxalic acid, ferrous ions, nitrite ion and iodides.
Oxidation states of Mn in \( MnO_2, KMnO_4 \) and \( K_2MnO_4 \) are
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → +4, +7, and +6 respectively
The oxidation state of an element in a compound is found from the rule that the oxidation states of all atoms add up to the charge on the species (zero for a neutral compound), taking O as −2 and K as +1. Let the oxidation state of Mn be x.
\( \displaystyle \ce{MnO2}: \quad x + 2(-2) = 0 \Rightarrow x = +4 \)
\( \displaystyle \ce{KMnO4}: \quad (+1) + x + 4(-2) = 0 \Rightarrow x = +7 \)
\( \displaystyle \ce{K2MnO4}: \quad 2(+1) + x + 4(-2) = 0 \Rightarrow x = +6 \)
So the oxidation states of Mn are +4, +7 and +6. This fits the passage: the green manganate ion (+6) disproportionates into permanganate (+7) and manganese dioxide (+4), one higher and one lower oxidation state.
Shortcut: \( \ce{K2MnO4} \) has one potassium more than \( \ce{KMnO4} \) with the same four oxygens, so its manganese must be one unit lower (+6 against +7). Only Option (4) gives +7 and then +6.
Note: It seems that there is a mistake in the original NTA paper. The second equation of the passage is printed as 3K2MnO4 + 4H+ → 2KMnO4 + MnO2 + 2H2O, in which the charges do not balance; it should be 3MnO42− + 4H+ → 2MnO4− + MnO2 + 2H2O.